Problem: Show that the axiom of choice (AOC) is equivalent to the statement that for any indexed family of non-empty sets, with , the product
Proof: This is pretty immediate when one writes down the actual definition of the product, namely:
So, if one assumes the AOC then one must assume the existence of a choice function
So, then if we consider as just a class of sets, the fact that we have indexed them implies there exists a surjective “indexing function$
where clearly since we have already indexed out set we have that . So, consider
This is clearly a well-defined mapping and and thus
from where it follows that
Conversely, let be a class of sets and let be an indexing function. We may then index by . Then, by assumption
Thus there exists some
Thus, we have that
is a well-defined mapping with
For each . It follows that we have produced a choice function for and the conclusion follows.
Remark: We have assumed the existence of a bijective indexing function , but this is either A) a matter for descriptive set theory or B) obvious since satisfies the conditions. This depends on your level of rigor.
Problem: Let be a set; let be an indexed family of spaces; and let be an indexed family of functions
a) Prove there is a unique coarsest topology on relative to whish each of the functions is continuous.
and let . Prove that is a subbasis for .
c) Show that the map is continuous relative to if and only if each map is continuous.
d) Let be defined by the equation
Let denote the subspace of of the product space . Prove taht the image under of each element of is an open set in .
a) We first prove a lemma
Lemma: Let be a topology on , then all the mappings are continuous if and only if where is defined in part b).
Proof:Suppose that all the mappings are continuous. Then, given any open set we have that is continuous and so is open and thus from where it follows that .
Conversely, suppose that . It suffices to prove that for a fixed but arbitrary . So, to do this let be open in then and thus by assumption ; but this precisely says that is open in . By prior comment the conclusion follows.
By previous problem is in fact a topology on , and by our lemma we also know that all the mappings are continuous since . To see that it’s the coarsest such topology let be a topology for which all of the are continuous. Then, by the other part of our lemma we know that and thus . So,
And thus is coarser than .
The uniqueness is immediate.
b) It follows from the previous problem that we must merely show that is a subbasis for the topology . The conclusion will follow from the following lemma (which was actually an earlier problem, but we reprove here for referential reasons):
Lemma: Let be a set and be a subbasis for a topology on . Then, the topology generated by equals the intersection of all topologies which contain .
Also, let be the topology generated by the subbasis .
Clearly since we have that .
Conversely, let . Then, by definition to show that it suffices to show that for a fixed but arbitrary . To do this we first note that by definition that
for some . Now, if we know (since ) that and thus
for each . It follows that is the union of sets in and thus . It follows from previous comment that .
The conclusion follows.
The actual problem follows immediately from this.
c) So, let be some mapping and suppose that is continuous for each . Then, given a subbasic open set in we have that
for some and for some open sets in respectively. Thus may be written as
but since each we see that is the finite union of open sets in and thus open in . It follows that is continuous.
Conversely, suppose that is continuous then is continuous since it’s the composition of continuous maps.
d) First note that
from where it follows that the initial topology under the class of maps on is the same as the initial topology given by the single map . So, in general we note that if is given the initial topology determined by then given an open set in we have that which is open in the subspace .
Problem: Suppose that for each the topology on is given by a basis . The collection of all sets of the form
Such that is a basis for with the box topology, denote this collection by . Also, the collection of all sets of the form
Where for finitely many and otherwise is a basis for the product topology on .
Proof: To prove the first part we let be open. Then, by construction of the box topology for each we may find some such that is open in and . So, then for each we may find some such that and thus
Noticing that and every element of is open finishes the argument.
Next, we let be open with respect to the product topology. Once again for each we may find some such that is open in for each and for all but finitely many , call them . So, for each we may find some such that and so
Noting that and is a collection of open subsets of finishes the argument.
Problem: Let be a collection of topological spaces such that is a subspace of for each . Then, is a subspace of if both are given the product or box topology.
Proof: Let denote the topologies inherits as a subspace of and as a product space respectively. Note that are generated by the bases (where is the basis on with the product topology), and
So, let where then
Where is open in , and thus is open in . Also, since for all but finitely many it follows that for all but finitely many . And so . Similarly, if then
Where is open in , but this means that for some open set in and so
From where it follows that and thus are equal.
The case for the box topology is completely analgous.
Problem: Let be a collection of Hausdorff spaces, then is Hausdorff with either the box or product topologies
Proof: It suffices to prove this for the product topology since the box topology is finer.
So, let be distinct. Then, for some . Now, since is Hausdorff there exists disjoint open neighborhoods of respectively. So, are disjoint open neighborhoods of respectively. The conclusion follows.
Problem: Prove that .
Clearly this is continuous since
Problem: One of the implications holds for theorem 19.6 even in the box topology, which is it?
Proof: If where the latter is given the box topology then we have that each is continuous and thus so is each . #
Problem: Let be a sequence of points in the product space . Prove that converges to if and only if the sequences coverge to for each . Is this fact true if one uses the box topology?
Proof: Suppose that is a neighborhood of such that
is infinite. Notice then that if that from where it follows that is a neighborhood of which does not contain all but finitely many values of contradicting the fact that in .
Conversely, suppose that for each and let be a basic open neighborhood of . Then, letting be the finitely many indices such that . Since each there exists some such that . So, let . Now, note that if then for some . But, since clearly we must have that and thus . It follows that for every we have that . Then, since every neighborhood of contains a basic open neighborhood and the above basic open neighborhood was arbitrary the conclusion follows.
Clearly the first part holds in the box topology since there was no explicit mention of the product topology or it’s defining characteristics. That said, the second does not hold. Consider . Clearly each coordinate converges to zero, but is a neighborhood of in the product topology. But, if one claimed that for every (for some that they’d be wrong. To see this merely note and so and thus .
Problem: Let be the subset of consisting of all eventually zero sequences. What is in the box and product topology?
Proof: We claim that in the product topology . To see this let be a basic non-empty open set in with the product topology. Since we are working with the product topology we know there are finitely many indices such that . So, for each select some and consider where
Clearly then and thus every non-empty open set in intersects and the conclusion follows.
Now, we claim that with the box topology that . To see this let . Then, there exists some subsequence of the sequence which is non-zero. For each form an interval such that . Then, consider
Clearly then is a neighborhood of and since each clearly has an infinite subsequence of non-zero values it is disjoint from . It follows that in with the box topology that is closed and thus as desired.
Problem: Given sequences and of real numbers with define by the equation
Show that if is given the product topology that is a homeomorphism. What happens if is given the box topology?
Proof: Let us first prove that is a bijection. To do this we prove something more general…
Lemma: Let and be two classes of untopologized sets and a collection of bijections . Then, if
we have that is a bijection.
Proof: To prove injectivity we note that if
And by definition of an -tuple this implies that
for each . But, since each is injective it follows that
For each . Thus,
To prove surjectivity we let be arbitrary. We then note that for each fixed we have there is some such that . So, if is the corresponding -tuple of these values we have that
from where surjectivity follows. Combining these two shows that is indeed a bijection.
Lemma: Let and be two classes of non-empty topological spaces and a corresponding class of continuous functions such that . Then, if and are given the product topologies the mapping
Proof: Since the codomain is a product space it suffices to show that
is continuous for each . We claim though that the diagram
commutes where and denote the canonical projections from and to and respectively. To see this we merely note that
which confirms the commutativity of the diagram. But, the conclusion follows since is the composition of two continuous maps (the projection being continuous since is a product space).
The lemma follows by previous comment.
We come to our last lemma before the actual conclusion of the problem.
Lemma: Let and be two classes of non-empty topological spaces and a set of homeomorphisms with . Then,
is a homeomorphism if and are given the product topology.
Proof: Our last two lemmas show that is bijective and continuous. To prove that it’s inverse is continuous we note that
And similarly for the other side. Thus,
Which is continuous since each is continuous and appealing to our last lemma again. Thus, is a bi-continuous bijection, or in other words a homeomorphism. The conclusion follows.
Thus, getting back to the actual problem we note that if we denote that each is a homeomorphism. Thus, since it is easy to see that
we may conclude by our last lemma (since we are assuming that we are giving in both the domain and codomain the product topology) that is a homeomorphism.
This is also continuous if we give the box topology. To see this we merely need to note that and thus if all of the are open then so are (since each is continuous) is each and thus the inverse image of a basic open set is the unrestricted product of open sets and thus basic open. A similar argument works for the inverse function.
Up until now we’ve discussed how the subspace topology reacts with subgroups and how the quotient topology interacts with the quotient group, it seems like a natural progression to then discuss how the product topology reacts with the direct product. So, as in the past we firstly need to verify that the two do agree, namely:
Theorem: Let be topological groups. Then, is a topological group with the direct product group structure and product topology.
Proof: Clearly is a set with both a group theoretic and topological structure and so it remains to show that the map
is continuous. But, note that
being the product of continuous maps is continuous. The conclusion immediately follows.
There are canonical topological epimorphisms from into . Namely:
Then, is an open topological epimorphism.
Proof: The fact that it’s open, surjective, and continuous follows since topologically is merely the canonical projection of a product space onto it’s th coordinate. Thus, it remains to show that it’s a homomorphism. But, this is a routine calculation.
Theorem: Let be as above. Then,
from where it follows from a previous theorem that
Now, so we can use it we should mention in passing that:
Theorem: where is any bijeciton.
In fact, that is all I wanted to mention about them. Unfortunately the interesting stuff about the direct product of topological groups that is interesting isn’t “accessible” and would take too much time to really discuss. Thus, for once I have a post which is less than one-thousand words. Enjoy.
In this post we will discuss one of the most important in point-set topology, Tychonoff’s theorem, and prove up some miscellaneous facts that are either interesting or useful.
Theorem (Tychonoff): Let be a non-empty class of non-empty compact topological spaces, then is compact under the product topology.
Proof: It follows from an earlier theorem that we must merely show that every class of closed sets in the defining closed subbase with the F.I.P have non-empty intersection. So, let be such a class. Consider some arbitrary , since is subbasic closed for ever we know that is closed, and so is a class of closed subsets of , and noticing that and so by assumption we have that . In other words, we have there exists some and so every element of contains an element which has as it’s th coordinate. Doing this for each produces a set such that . In light of our initial comments the conclusion follows.
From this we can prove a nice little result that is used incessantly in multivariate analysis. But, before we continue we need a little terminology
n-cell: An -cell is a multi-dimensional generalization of a closed interval in . In other words, we call an -cell if for every . In other words, is an -cell if there exists some such that .
Theorem (Generalized Heine-Borel): Every closed and bounded subspace of is compact.
Proof: Let be such a set. Since is bounded we know that is contained in some -cell . So, if we prove that every -cell is compact we are done (since a closed subspace of a compact space is compact).
But, we know that and since is compact we have by Tychonoff’s theorem that the product of compact spaces is compact. The conclusion follows.
Be careful about taking this and attempting to try to make a generalization of the extreme value theorem, namely “$If is continuous and compact then “. It fails for an easily over-looked and decidedly non-topological property of . Particularly, we have that has a canonical ordering (the one we know and love), but what is the ordering on ? No ordering no concept of infimums and supremums.
We now take a look at some of the odds and ends that were previously mentioned. It should probably be mentioned for those that are reading this and already know topology that, for now, we omit mention of the majority (save Hausdorfness AKA ) of the separability axioms and things like local compacntess, connectedness, paracompactness, and path-connectedness. We shall mention them in due time, from where we will discuss which are invariant under product.
Theorem: Let be a finite number of discrete spaces, then the product set is the same topological space under both the product and discrete topology.
Proof: Let be the product topology on and, of course, let be the discrete topology (I use the for power set). It is clear that and so we must merely show the reverse inclusion to finish.
So, let then for some subsets , but since is discrete we know that each is open. Thus, is the finite product of open sets and thus open in the product (box) topology. The conclusion follows.
The fact that I noted that product topology and box topology were the same here was meaningful. We know that the two coincide for the product of finitely many spaces, but the next remark shows (once again) how they differ for infinite.
Theorem: Let be an arbitrary collection of discrete spaces, then under the box topology coincides with under the discrete topology. The same cannot be said about the product topology.
Proof: Follows the exact same reasoning as in the last theorem.
To see why the infinite product of discrete spaces need not be discrete under the product topology, one need only consider where under the discrete topology. Clearly then is open in under the discrete topology but NOT the product topology.
The last thing we will talk about is the product of maps.
Product of maps: Let and be two maps, then define the product of the maps to be by .
We extend this definition to an arbitrary number of topological spaces as follows: Let and be two sets of topological spaces and let be a class of mappings such that (this can obviously be formulated in the guise of ordered triples as well), we then define by
The reader may recall that this concept was used to prove that if is continuous and Hausdorff then is a closed subset of .
We begin by proving the product of continuous maps is continuous. This is the first of a sequence of three proofs which will lead to a very satisfying corollary.
Theorem: Let and be an arbitrary collection of topological spaces and let be a corresponding set of continuous mappings such that . Then, is continuous, where are under the product topology.
Proof: Let then, which clearly means
that and so and thus
Conversely, let . Then, and thus and thus . The conclusion follows.
Now, as was previously proven we need only prove that the inverse image of each open basic set is open. So, let (the defining open base for ). Then, where for all but finitely many , and thus . But, we see that since is an open set which equals for all but finitely many ‘s that is going to be an open set (since each is continuous) which equals for all but finitely many ‘s. The conclusion follows.
Theorem: Let everything be as in the previous theorem, except instead of each being continuous, let it be bijective. Then is bijective.
Injectivity: Suppose that . Then, by definition and since two functions (remember we think of arbitrary tuples as functions) are equal if and only if they take the same value at every point we see that and since each is bijective (and thus injective) it follows that , from where injectivity follows.
Surjectivity: Let be arbitrary. For each we have that there exists some such that and thus from where surjectivity immediately follows.
Theorem: Let everything be as before, except now instead of continuous or bijective assume that each is surjective and open. Then, is open.
Proof: Let then for some . Clearly then we have that and so and so .
Conversely, let then for some and thus
The conclusion follows. .
Now, let be open in then where for all but finitely many it follows then that is an open set which equals for all but finitely many (since each is surjective and open). Thus which by prior comment is the product of open sets in which equals the full space for all but finitely many . It follows that is open in under the product topology.
Theorem: Combining these three we see that if for all , then there exists some homeomorphism (a bijective continuous open map) and so is a bijective continuous open map and thus a homeomorphism. It follows that where each are under the product topology.
In the previous post we gave the definitions of the product topology and some nice little tools about it. Here, we give some theorems regarding what properties of topological spaces are invariant under the formation of the product topology. Let us start out nice and easy
Theorem: Let be a finite number of topological spaces with corresponding open bases . Then, is an open base for under the product topology.
Proof: Let be open in , and let . There exists some (the defining open base) such that . But, by theorem 1 in the last post we know that is an open set containing for and since is an open base for we know there exists some such that . Clearly then and since and were arbitrary the conclusion follows.
We know use this to prove a nice little theorem.
Theorem: Let be a finite number of second countable spaces, then under the product topology is second countable.
Proof: Let be as above. Since is an open base for is remains to prove that it is countable. Let be given by . It is clear that this is a bijection. Recalling that the finite product of countable sets is countable finishes the argument.
It turns out that the above is a little more restrictive than is necessary as the following theorem illustrates.
Theorem: Let be a sequence of second countable spaces, then is second countable under the product topology.
Proof: Let be the corresponding countable open bases and define where and .
It is relatively easy to prove that given by is a bijection. It follows then, since the finite product of countable sets is countable, that is countable.
So, now define . It is clear that since is the countable union of countable sets that it is countable. So now, it remains to prove that is in fact an open base for
So, let be open in and let . There exists some in the defining open base such that . Considering the form in which basic open sets in the product topology take we know that there is a finite set of indices such that . So, let . Clearly we have that is a neighborhood of for and since is an open base there exists some such that . So, let . Then, it is evident that . Noting that finishes the argument. .
Note the necessity for finiteness again.
Theorem: Let be a finite number of toplogical spaces and let be a number of corresponding dense sets. Then, is dense in under the product topology.
Proof: Let be arbitrary and let be any neighborhood of . We then have by theorem 1 in the last post that is a neighborhood of for and since is dense in there exists some such that . Clearly then , and clearly . Noting the arbitrariness of finishes the argument.
From this we can derive the obvious consequence.
Theorem: Let be finite number of separable spaces and let under the product topology, then is separable.
Proof: Since each is separable it has a countable dense subset and by the above theorem we see that is dense in , but since the finite product of countable sets is countable the conclusion follows.
As in the case of second countable spaces we may extend this to the countable case
Theorem: Let be sequence of separable spaces, then is separable.
Proof: Since each is separable it has a countable dense subset we may extract an arbitrary but fixed sequence from them as where . We now define where
Clearly we have that by is a bijection and since the finite product of countable sets is countable it follows that is countable. So, let being the countable union of countable sets is countable.
From the previous paragraph it is clear that we must merely show that is dense in . To do this let be arbitrary and any neighborhood of . There exists some (the defining open base) such that . But, we know that for finitely man indices, say . Let . We know then that is a neighborhood of for every and since is dense in there exists some such that . Clearly then, we have that where is in and thus . But, and the conclusion follows.
We now discuss the product of Hausdorff spaces.
Theorem: be an arbitrary collection of Hausdorff spaces, then under the product topology is Hausdorff.
Proof: Let such that then for some and since is Hausdorff we know that there exists open sets such latex and . So define where and define where clearly we have that and and since we must have . The conclusion follows.
NOTE: All sets and spaces in the following discussion will be assumed to be, without loss of generality, non-empty. Also, if not stated it is assumed that under the product topology where is a non-empty collection of topological spaces.
To continue any further we mus make an, admittedly not brief, sojourn into the concept of product topologies. Now, this isn’t a text-book so I’ll assume that the reader is familiar with the basic concepts of Cartesian products and projections (denote ). With this in mind we first make a definition for the arbitrary Cartesian product of sets:
Arbitrary Cartesian Product: Let be an arbitrary class of non-empty sets, then we define .
Now, if is a non-empty class of topological spaces we define the product topology on as the topology generated by the subbase (where is the topology on ,) in the sense that it is the set of all finite intersections of elements of form an open base for , and thus the topology on is made up of all arbitrary unions of elements of this open base. We call in the above discussion the defining open subbase for and the defining open base for under the product topology.
So what do elements of look like? Well, let be an open set in then is (in words) all such that . It is clear that the only limitation then that needs to be made on is that it’s th coordinate is in . From this it is clear that where . Thus, is made up of all sets of the form where for all but one
Example: If under the usual topology then whereas
So, this being said we see that the elements of (being the finite intersection of elements of ) are going to be the product of open sets such that for all but finitely many .
So, let’s start proving some theorems!
Theorem 1: All projections are open mappings.
Proof: Let be open then where , and so and since the th coordinate of every basic open set (whether equal to the full space or not) is an open set in and so is the union of open sets in and thus open.
Corollary 1: If is a neighborhood of then is a neighborhood of
Remark: Projections need not be closed mappings. Consider the set as a subset of with the product topology. It is relatively easy is closed, but which is open.
One might ask “Why not just make the open base for a topology on the set of all arbitrary products of open sets?”. The topology that this is describing is called the box topology. Namely, the box topology has the set of all arbitrary products of open open sets as an open base. It is clear that the box and product topologies coincide for finite cases but do not for infinite cases. So, why did we choose the box topology over the product topology? The answer comes from the following theorem.
Theorem: Let be a mapping of a topological space into the product space . Then, is continuous if and only if is continuous for all .
Let be open in then and since is subbasic open in and continuous we see that is open in .
Lemma: A mapping (where and are arbitrary topological spaces) is continuous if and only if is open for every where is an open base for .
: Let be open in then where . So then, which by assumption is the union of open sets in and thus open. The conclusion follows.
: This is obvious.
So, suppose that is basic open in , then where and is open in . It follows then that
and since each is continuous it follows that is the finite intersection of open sets in and thus open. The conclusion follows.
Note that the key point here was that was the finite intersection of inverse images of open sets. If we had the box topology we could have had that was the infinite intersection of inverse images of open sets we would have no guarantee that would be open since it would be the infinite intersection of open sets…which we all know need not be open.
Now, one may ask “Topologically does it matter which order we product the sets?”. The answer is “no…for finite cases”
Theorem: Let be a finite collection of topological spaces then where is an permutation of
Proof: It is readily verified that the canonical mapping given by is a homeomorphism.